What is the series generated by the generating function : $\frac{1}{(1-x)^2}$?

Question

I know that $$\frac{1}{(1-x)} = x^0 + x^1 + x^2 + x^3 + \space ... \space\space\space\space\space\space\space\space (1)$$

When I differentiate $(1)$ with respect to $x$ I get the following function : $$\frac{x}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 + \space ...\space\space\space\space\space (2)$$

If I divide LHS and RHS of $(2)$ by $x$ I would get :

$$\frac{1}{(1-x)^2} = \frac{1}{x} + 2 + 3x + 4x^2 + \space...$$

But from what I have seen from here my derivation of the function $ \frac{1}{(1-x)^2}$ is completely wrong.

What is it that I am doing wrong here?

Answer 1

Try redoing the derivative of $1/(1-x)$ more carefully.

Answer 2

$$\frac{d}{dx}\frac1{1-x}=\frac1{(1-x)^2}.$$

Answer 3

A different approach, by a manual verification,

$$ \frac{1}{1-x} = \sum_{n \geq 0}X^n, $$

and so

$$ \frac{1}{(1-x)^2} = \frac{1}{1-x} \cdot \frac{1}{1-x} = \sum_{n \geq 0}X^n \cdot \sum_{m \geq 0}X^m = \sum_{i \geq 0}\left(\sum_{k = 0}^i1\right)X^i = \sum_{i \geq 0}(i+1)X^i $$